Question

Let X denote the number of bars of service on your cell phone whenever you are at an intersection with the following probabilities.
x 0 1 2 3 4 5
P(X=x) 0.1 0.15 0.25 0.25 0.15 0.1

Determine the following probabilities
a. Two or three bars
b. At least one bar
c. Fewer than two bars
d. More than three bars

Answer 1

Answer:

a. $P(2 \leq X \leq 3) = 0.5$

b. $P(X \geq 1) = 0.9$

c. $P(X < 2) = 0.25$

d. $P(X > 3) = 0.25$

Step-by-step explanation:

We are given the following distribution:

$P(X = 0) = 0.1$

$P(X = 1) = 0.15$

$P(X = 2) = 0.25$

$P(X = 3) = 0.25$

$P(X = 4) = 0.15$

$P(X = 5) = 0.1$

a. Two or three bars

$P(2 \leq X \leq 3) = P(X = 2) + P(X = 3) = 0.25 + 0.25 = 0.5$

Thus:

$P(2 \leq X \leq 3) = 0.5$

b. At least one bar

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.1 = 0.9$

Thus:

$P(X \geq 1) = 0.9$

c. Fewer than two bars

$P(X < 2) = P(X = 0) + P(X = 1) = 0.1 + 0.15 = 0.25$

Thus:

$P(X < 2) = 0.25$

d. More than three bars

$P(X > 3) = P(X = 4) + P(X = 5) = 0.15 + 0.1 = 0.25$

Thus:

$P(X > 3) = 0.25$